Simplify and expand the following expression: $ \dfrac{3}{4k + 9}-\dfrac{k - 9}{5k - 9} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(4k + 9)(5k - 9)$ Multiply the first term by $\dfrac{5k - 9}{5k - 9}$ $ \begin{align*} \dfrac{3}{4k + 9} \times \dfrac{5k - 9}{5k - 9} & = \dfrac{(3)(5k - 9)}{(4k + 9)(5k - 9)} \\ & = \dfrac{15k - 27}{(4k + 9)(5k - 9)}\end{align*} $ Multiply the second term by $\dfrac{4k + 9}{4k + 9}$ $ \begin{align*} \dfrac{k - 9}{5k - 9} \times \dfrac{4k + 9}{4k + 9} & = \dfrac{(k - 9)(4k + 9)}{(5k - 9)(4k + 9)} \\ & = \dfrac{4k^2 - 27k - 81}{(5k - 9)(4k + 9)}\end{align*} $ Now we have: $ = \dfrac{15k - 27}{(4k + 9)(5k - 9)} - \dfrac{4k^2 - 27k - 81}{(5k - 9)(4k + 9)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{15k - 27 - (4k^2 - 27k - 81)}{(4k + 9)(5k - 9)} $ $ = \dfrac{15k - 27 - 4k^2 + 27k + 81}{(4k + 9)(5k - 9)} $ $ = \dfrac{42k + 54 - 4k^2}{(4k + 9)(5k - 9)}$ Expand the denominator: $ = \dfrac{42k + 54 - 4k^2}{20k^2 + 9k - 81}$